I am unsure on how to solve this problem. The only clue i have is to integrate ROC to find time.
A jet aircraft has the capacity to fly at an absolute altitude ceiling of 11000 m, and a rate of climb (ROC) at sea level of 20.87 m/s. Calculate the time to climb from sea level to 8000 m. Assume a linear variation of the rate of climb with altitude during the whole maneuver.
The fastest way is an integration. If you plot the inverted climb speed over altitude like in the plot below, the area under the curve will give you the time to climb. I would suggest you calculate the climb speed at every 1000 m by rule of proportion — at sea level it is 20.87 m/s and at 11.000 m it is 0 m/s. Then you get a plot like the one below. The red circles are the calculated points; the lines between them are straight interpolations.
Since the unit on the Y-axis is seconds per meter and the unit on the X-axis is meters, the area is in seconds. Just calculate the area of each trapezoid between the start and the end altitudes and add them. In the figure below I have crosshatched the trapezoids between 3000 m and 8000 m.
Inverted climb speed over altitude plot. Caution — this is a generic plot I did for this purpose earlier, so it uses different numbers. I crosshatched the area between 3000 m and 8000 m, because I made the plot for calculating the climb time from 3000 m to 8000 m.
In general, you need to integrate. But in this particular problem, the rate of climb goes down linearly, so the integration is trivial. At sea level the rate of climb is 20.87 m/s. At 11,000 ft. It's 0. So it goes down by 20.87/11 m/s (1.90) for each thousand feet of climb. The rate of climb at 8000 ft. would be 20.87 - 8*1.90, which is 5.6 m/s. The other nice thing about it being linear is that you can use the average rate of climb. That's (20.87 + 5.6) / 2, which is 13.24. At 13.24 m/s, climbing 8000 m will take 8000/13.24, which is 604 seconds, or just over 10 minutes.
